Answer
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Hint: We know that at the maximum height, the velocity is minimum. We will use one of the equations of motion and substitute the required values. We will manipulate accordingly and obtain the result.
Complete step by step solution:
Given, The maximum height achieved by Kangaroo, \[H = 2.62\]. Final velocity at the maximum height is, \[v = 0\]. Acceleration due to gravity is, \[g = - 9.8\,{\text{m }}{{\text{s}}^{ - 2}}\]. Here, minus is for downward direction.
We know that,
${v^2} - {u^2} = 2gH \\
\Rightarrow 0 - {u^2} = 2\left( { - 9.8} \right) \times \left( {2.62} \right) \\
\Rightarrow u = \sqrt {2 \times \left( {9.8} \right) \times \left( {2.62} \right)} \\
\Rightarrow u = 7.17\,{\text{m }}{{\text{s}}^{ - 1}} \\$
Hence, the required answer is \[7.17\,{\text{m }}{{\text{s}}^{ - 1}}\]. The correct option is D.
Additional information:
Equation of motion: In physics, equations of motion are equations that define a physical system's action as a function of time in terms of its motion. More precisely, in terms of dynamic variables, the equations of motion define a physical system's action as a series of mathematical functions. Typically, these variables are spatial coordinates and time, but can include components of momentum. Generalized coordinates are the most common choice, which can be any convenient variables that describe the physical system. In classical mechanics, functions are described in Euclidean space, but are replaced by curved spaces in relativity. If a system's dynamics are known, the equations are the solutions to the differential equations that explain the dynamics' motion.
Note: The motion equations of kinematics define the most basic principles of an object's motion. In \[1{\text{D}}\], \[{\text{2D}}\] and \[{\text{3D}}\], these equations govern the motion of an object. They can easily be used at different times to measure expressions such as an object's position, velocity, or acceleration. Most of the students tend to make mistakes by not considering the sign convention of acceleration due to gravity. When a body moves up, it is going against gravity.
Complete step by step solution:
Given, The maximum height achieved by Kangaroo, \[H = 2.62\]. Final velocity at the maximum height is, \[v = 0\]. Acceleration due to gravity is, \[g = - 9.8\,{\text{m }}{{\text{s}}^{ - 2}}\]. Here, minus is for downward direction.
We know that,
${v^2} - {u^2} = 2gH \\
\Rightarrow 0 - {u^2} = 2\left( { - 9.8} \right) \times \left( {2.62} \right) \\
\Rightarrow u = \sqrt {2 \times \left( {9.8} \right) \times \left( {2.62} \right)} \\
\Rightarrow u = 7.17\,{\text{m }}{{\text{s}}^{ - 1}} \\$
Hence, the required answer is \[7.17\,{\text{m }}{{\text{s}}^{ - 1}}\]. The correct option is D.
Additional information:
Equation of motion: In physics, equations of motion are equations that define a physical system's action as a function of time in terms of its motion. More precisely, in terms of dynamic variables, the equations of motion define a physical system's action as a series of mathematical functions. Typically, these variables are spatial coordinates and time, but can include components of momentum. Generalized coordinates are the most common choice, which can be any convenient variables that describe the physical system. In classical mechanics, functions are described in Euclidean space, but are replaced by curved spaces in relativity. If a system's dynamics are known, the equations are the solutions to the differential equations that explain the dynamics' motion.
Note: The motion equations of kinematics define the most basic principles of an object's motion. In \[1{\text{D}}\], \[{\text{2D}}\] and \[{\text{3D}}\], these equations govern the motion of an object. They can easily be used at different times to measure expressions such as an object's position, velocity, or acceleration. Most of the students tend to make mistakes by not considering the sign convention of acceleration due to gravity. When a body moves up, it is going against gravity.
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